I think I'm fundamentally misunderstanding something.

Say I have a gauged Lagrangian for a complex scalar field $\phi$ with no SSB:

$$\begin{equation} \mathcal{L} = (D_{\mu}\phi)(D^{\mu}\phi)^{\dagger} - m^2 \phi \phi^{\dagger} - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}\tag{1} \end{equation}$$

with $D_{\mu} = \partial_{\mu} + i A_{\mu}$.

And now suppose I parameterise my complex scalar field as $\phi(x) = r(x) e^{i \theta(x)}$ -- two real degrees of freedom excited around the vacuum at $\langle \phi\rangle = 0$. If I now plug this into the Lagrangian I get

$$\mathcal{L} = (\partial_{\mu} r + i \partial_{\mu} \theta r + i A_{\mu} r )(\partial_{\mu} r - i \partial_{\mu} \theta r - i A_{\mu} r ) - m^2 r^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}.$$

But by gauge invariance $A_{\mu}$ and $A_{\mu} + \partial_{\mu} \theta$ are exactly the same field (this may be the place where I'm doing something wrong), so

$$\mathcal{L} = (D_{\mu}r)(D^{\mu}r)^{\dagger} - m^2 r^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu},\tag{2}$$

which only depends on the real excitation!

I'm very confused as to where the angular excitation has gone. Was it just never real in the first place?

If we had SSB, we'd happily eliminate the goldstones that corresponded to the gauge degrees of freedom by letting them get eaten by the gauge field in just such a way. The only difference here is there's no vev to give the $A$s a mass.

In this post, TwoBs' answer seems to do the same as I do, but it seems to me their argument rests on the fact that they have neglected to package up (where $h$ corresponds to my $r$) $\partial_{\mu} h$ and $A_{\mu} h$ into a covariant derivative again, and they claim this makes $\mathcal{L}$ non-gauge-invariant. I don't understand the argument. I don't feel like I have fixed a gauge anywhere, I've just expressed the $\phi$ field in a certain form, and since $A_{\mu}$ was a general field, $A_{\mu} + \partial_{\mu} \theta$ is surely also general.

Is there a difference between a gauge transformation and an angular excitation of the $\phi$ field? Do they only look the same at the level of the Lagrangian and in reality they're truly different?

  • $\begingroup$ I think what you wrote is correct and expresses the notion that $\mathcal{L}$ has a global $U(1)$ symmetry. $\endgroup$
    – proton
    Commented Jul 20, 2020 at 10:33
  • 1
    $\begingroup$ (a) The kinetic term for $\theta$ vanishes at $r=0$, so the degree-of-freedom counting can go awry. (b) If $r>0$, your statement is just that the gaueg transformation can be used to make the field real, and that the "phase" is the goldstone mode. $\endgroup$
    – Toffomat
    Commented Jul 20, 2020 at 11:47
  • $\begingroup$ I'm sorry, I still don't follow. I haven't tried to 'make' the field real, I've just expressed it in a normal way for complex numbers and the real part is the only one which survived. Is it still called a goldstone mode if there's no broken symmetry? I'm clearly just not making a mental connection here :( $\endgroup$
    – quixot
    Commented Jul 20, 2020 at 12:43
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    $\begingroup$ Sorry for the terse comment (I'm in a hurry at the moment), but here's a relevant detail: $\partial_\mu r$ is undefined at $r=0$. The paper Unitary gauge considered harmful is also related (and entertaining), although I'm not sure it answers your question. $\endgroup$ Commented Jul 20, 2020 at 14:30
  • $\begingroup$ To second @Toffomat's point: switch off the gauge field for a second (set the charge =0). You have two massive degrees of freedom, not a massive and a massless one, just because you don't see a mass term for the phase (not Goldstone mode, but a notional/aspirational fixing-to-be-goldston one) in your polar representation. $\endgroup$ Commented Jul 20, 2020 at 15:59

2 Answers 2


Indeed you're doing something wrong, or at least confusing, when you say "...by gauge invariance $A_\mu$ and $A_\mu+\partial_\mu \theta$ are exactly the same field..." and then discard the $\partial_\mu \theta$: You're implictly imposing a gauge transformation with parameter $\theta$, which absorbs the phase into the gaueg field.

Also, your decomposition is only valid for nonzero fields, which includes extra confusion. For example, the modulus $r$ should be gauge invariant, you should have two massive scalars etc.

When there is a nonzero VEV, all you have writen goes through, and it's essentially the standard Higgs mechanism, where you can and routinely do shuffle "angular excitaion of $\phi$" and "part of gauge field" around by treating $\theta$ as parameter of a gauge transformation.

  • $\begingroup$ Right, so the gauge invariance is the statement that I'm allowed to push degrees of freedom from the $\phi$ field into the gauge field? I feel like that should be reflected in the non-SSB case as well, but what kind of dof is it for the gauge field now? Massive? But surely you need SSB for that. $\endgroup$
    – quixot
    Commented Jul 20, 2020 at 17:38
  • $\begingroup$ @quixot Well, rewriting doesn't change dofs, so you still have two massive scalar dofs and two massless vector ones. The fact that that's not apparent is due to the ill-defined decomposition. And your first question: gauge invarince means that you can perform a transformation that changes the gauge field and the charged matter fields. Whether that "pushes dofs around" in a simple manner depends on the situation (such as ssb or no ssb) $\endgroup$
    – Toffomat
    Commented Jul 20, 2020 at 17:52
  • $\begingroup$ Sure I see, of course it might not be nice-looking or simple (and might be ill-defined in some scenarios). So if I understand correctly, the special thing about after SSB is that you can choose your decomposition along flat vs massive directions, and these are real, physical dofs and massless - but they mix with the gauge bosons and so when you mass-diagonalise you push both the old boson and the 'goldstone' into a new dof with a mass. But without SSB pushing one of the dofs into a boson doesn't correspond to any mass eigenstate? Or are you not able to do it at all? $\endgroup$
    – quixot
    Commented Jul 20, 2020 at 18:08
  • $\begingroup$ @quixot In general, you can't do it at all. $\endgroup$
    – Toffomat
    Commented Jul 21, 2020 at 7:28

The issue is that the scalar has to be transformed as well when you perform a gauge transformation. Namely if you want to perform the change $$A_\mu + \partial_\mu \theta(x)$$ it has to come along with a gauge transformation of the scalar field by the same parameter, i.e. $$\phi \rightarrow e^{-i\theta(x)}\phi$$ I would recommend writing the complete Lagrangian first with whatever parameterization for $\phi$ you want, for example $\phi = \rho(x)e^{i\sigma(x)}$, (notice that so far the parameters have nothing to do with a gauge transformation). One does not need to speak about VEV even, to perform a gauge transformation. This way you will explicitly see if you can pick some $\theta(x)$ such that it eliminates the $\sigma(x)$ field.

If I am not mistaken what will happen is that $\sigma(x)$ will decouple for an appropriate $\theta(x)$ meaning it will not have interaction terms, but the $\sigma(x)$ kinetic term should remain and the total number of degrees of freedom conserved.


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